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levacccp [35]
3 years ago
14

Vector G is 40.3 m long in a -35 degree direction. Vector H is 63.3 m long in a 270 degree direction. Find the magnitude of thei

r vector sum.
Physics
2 answers:
KIM [24]3 years ago
8 0

Answer:

The easiest way to tackle this problem is by breaking the vectors up into their components (also it's easier to type this method).

G = 40.3 cos (-35.0°) i + 40.3 sin (-35.0°) j ----where i denotes the x direction and j denotes the y direction

H = 63.3 sin (270°) j ----- I left off the i component because i know cos 270 is 0.

40.3 and 63.3 are the magnitudes of your vector, or the hypotenuse of a triangle. We are finding the horizontal (Adjacent) and vertical (Opposite) legs the respective hypotenuse (we have to remember SOH COH TOA)

Now you can simply add components

G + H = (40.3 cos (-35.0)) i + (40.3 sin (-35.0°)+ 63.3 sin (270)) j ----be sure to pay close attention to signs when doing your calculations on your calculator.

andrey2020 [161]3 years ago
3 0

Answer:

answer 30000000000000000000000

Explanation:

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Answer:

\partial \theta = 0.003

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we know that

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\theta = 1.484°

\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians

as we see that sin\theta = \theta

relative error\frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}

Where X_1 IS HEIGHT OF ROCK

X_2 IS THE HEIGHT OF ROAD

\partial X = uncertainity in measuring  distance

\partial X = 0.05

Putting all value to get uncertainity in angle

\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}

solving for \partial \theta we get

\partial \theta = 0.003

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